8.1 Preparing Alkenes: A Preview of Elimination Reactions

Elimination Reactions to Prepare Alkenes

Types of Alkene Elimination Reactions

Alkenes are formed through elimination reactions, where atoms or groups are removed from adjacent carbons to create a carbon-carbon double bond. Two main types show up repeatedly in organic chemistry:

Dehydrohalogenation removes a hydrogen halide (HX) from an alkyl halide to form an alkene.

• Requires a strong base such as sodium ethoxide ($\text{NaOCH}_2\text{CH}_3$) or potassium tert-butoxide ($\text{KOC(CH}_3\text{)}_3$)
• General reaction: $\text{R–CH}_2\text{–CH}_2\text{–X} \rightarrow \text{R–CH=CH}_2 + \text{HX}$

Dehydration eliminates water ($\text{H}_2\text{O}$) from an alcohol to produce an alkene.

• Uses a strong acid catalyst, typically sulfuric acid ($\text{H}_2\text{SO}_4$) or phosphoric acid ($\text{H}_3\text{PO}_4$)
• Requires high temperatures (typically 100–200°C)
• General reaction: $\text{R–CH}_2\text{–CH}_2\text{–OH} \rightarrow \text{R–CH=CH}_2 + \text{H}_2\text{O}$

Notice the pattern: both reactions pull off atoms from two adjacent carbons (the $\alpha$ and $\beta$ carbons) to generate the double bond. The difference is what's being removed and what drives the removal.

Dehydrohalogenation and Dehydration Processes

Dehydrohalogenation works because a strong base abstracts a proton from the carbon adjacent to the one bearing the halide (the $\beta$-carbon). As the proton leaves, the halide departs as a leaving group, and the electrons form the new $\pi$ bond.

• Example: $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{NaOCH}_2\text{CH}_3 \rightarrow \text{CH}_3\text{CH=CH}_2 + \text{NaBr} + \text{HOCH}_2\text{CH}_3$

Dehydration follows a different sequence. The acid first protonates the hydroxyl group ($\text{–OH}$), converting it into $\text{–OH}_2^+$, which is a much better leaving group than $\text{OH}^-$. Water then departs, and a proton is lost from the adjacent carbon to form the double bond.

• Example: $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{H}_2\text{SO}_4, \Delta} \text{CH}_3\text{CH=CH}_2 + \text{H}_2\text{O}$

The key takeaway: dehydrohalogenation is base-promoted, while dehydration is acid-catalyzed. Mixing these up is a common mistake on exams.

Predicting Alkene Products and Isomers

When more than one elimination product is possible, you need a way to predict which one dominates. That's where Zaitsev's rule comes in: the more substituted alkene is typically the major product. More substituted double bonds are more stable due to hyperconjugation (overlap of adjacent $\text{C–H}$ or $\text{C–C}$ $\sigma$ bonds with the $\pi$ system).

Hofmann elimination is the exception. Less substituted alkenes become the major product when you use a bulky base (like potassium tert-butoxide) or when the leaving group is poor (like a quaternary ammonium salt). The bulky base can't easily reach the more substituted, more sterically hindered $\beta$-hydrogen, so it grabs the more accessible one instead.

Elimination reactions can also produce different types of isomers:

• Constitutional (structural) isomers differ in where the double bond is located. For example, $\text{CH}_3\text{CH=CHCH}_3$ (but-2-ene) and $\text{CH}_2\text{=CHCH}_2\text{CH}_3$ (but-1-ene) are constitutional isomers.
• Cis/trans (geometric) isomers differ in the spatial arrangement of substituents across the double bond. Cis places substituents on the same side; trans places them on opposite sides. Trans alkenes are generally more stable due to reduced steric strain.

Steps to predict products:

1. Identify the precursor (alcohol or alkyl halide) and locate the carbon bearing the leaving group (the $\alpha$-carbon).
2. Find all $\beta$-carbons (those adjacent to the $\alpha$-carbon) that carry at least one hydrogen.
3. Draw the alkene that would result from eliminating toward each $\beta$-carbon.
4. Apply Zaitsev's rule to identify the major product (most substituted alkene).
5. Check whether cis/trans isomers are possible for any of those products.

Mechanistic Considerations in Elimination Reactions

Elimination reactions proceed by one of two mechanisms, and knowing which one applies changes your predictions.

E2 (bimolecular elimination) occurs in a single concerted step. The base removes a $\beta$-proton at the same time the leaving group departs, and the double bond forms simultaneously. The rate depends on the concentration of both the substrate and the base: $\text{rate} = k[\text{substrate}][\text{base}]$.

• Favored by strong bases, primary or secondary substrates, and higher base concentration
• Requires anti-periplanar geometry: the $\text{H}$ and the leaving group must be on opposite sides of the $\text{C–C}$ bond (180° dihedral angle) for the orbitals to align properly
• This geometric requirement means E2 reactions are stereospecific, often giving one stereoisomer preferentially

E1 (unimolecular elimination) occurs in two steps. First, the leaving group departs to form a carbocation intermediate (slow, rate-determining step). Then a base removes a proton from a carbon adjacent to the carbocation to form the alkene (fast step). The rate depends only on substrate concentration: $\text{rate} = k[\text{substrate}]$.

• Favored by tertiary substrates, weak bases, and polar protic solvents
• Because a planar carbocation intermediate forms, E1 reactions typically produce a mixture of stereoisomers (both cis and trans)
• Carbocation rearrangements (hydride or methyl shifts) are possible, which can lead to unexpected products

At this stage, recognizing which mechanism applies and how it affects the product distribution is the main goal. You'll explore E1 and E2 in much greater depth in later sections.